Problem: Let $y=\log_4(x^3-7x+5)$. Find $\dfrac{dy}{dx}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{3x^2-7}{x^3-7x+5}$ (Choice B) B $\dfrac{3x^2-7}{\ln(4)(x^3-7x+5)}$ (Choice C) C $\dfrac{1}{\ln(4)(x^3-7x+5)}$ (Choice D) D $\dfrac{\ln(4)(3x^2-7)}{x^3-7x+5}$
$\log_4(x^3-7x+5)$ is a logarithmic function, but its argument isn't simply $x$. Therefore, it's a composite function. In other words, suppose $u(x)=x^3-7x+5$, then $y=\log_4\Bigl(u(x)\Bigr)$. $\dfrac{dy}{dx}$ can be found using the following identity: $\dfrac{d}{dx}\left[\log_4\Bigl(u(x)\Bigr)\right]=\dfrac{u'(x)}{\ln(4)u(x)}$ [Why is this identity true?] Let's differentiate! $\begin{aligned} &\phantom{=}\dfrac{dy}{dx} \\\\ &=\dfrac{d}{dx}\log_4(x^3-7x+5) \\\\ &=\dfrac{d}{dx}\log_4\Bigl(u(x)\Bigr)&&\gray{\text{Let }u(x)=x^3-7x+5} \\\\ &=\dfrac{u'(x)}{\ln(4)u(x)} \\\\ &=\dfrac{3x^2-7}{\ln(4)(x^3-7x+5)}&&\gray{\text{Substitute }u(x)\text{ back}} \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=\dfrac{3x^2-7}{\ln(4)(x^3-7x+5)}$.